Tugas Termodinamika Demas Ganteng [PDF]

Citation preview

Adhemaz Noercahya Putranto 2020710450217 TUGAS TERMODINAMIKA 10 April 2021



1. Calculate the reversible work done in compressing 0.0283 m3 of mercury at a constant temperature of 273.15 K(O°C) from 1 atm to 3000 atm. The isothermal compressibility of mercury at 273.15 K(O"C) is where P is in atm and K is in atm-'.



2. Calculate Z and V for ethylene at 298.15 K (25°C) and 12 bar by the following equations: (a) The truncated virial equation [Eq. (3.39)] with the following experimental values of virial coefficients: (b) The truncated virial equation [Eq. (3.37)], with a value of B from the generalized Pitzer correlation [Eq. (3.59)]. (c) The RedlicWKwong equation. (d) The Soave/Redlich/Kwong equation. (e) The PengJRobinson equation.



3. Calculate the molar volume of saturated liquid and the molar volume of saturated vapor by the RedlichIKwong equation for one of the following and compare results with values found by suitable generalized correlations. (a) Propane at 313.15 K(40°C) where PSat = 13.71 bar. (b) Propane at 323.15 K(50°C) where PSat = 17.16 bar. (c) n-Butane at 373.15 K(100oC) where PSat = 15.41 bar.



Jawaban



a. Dik : v = 0,0283 m3 = 1 ft3 T = 273,15 K P1 = 1 atm P2 = 3000 atm



κ = 3,9 x 10-6 – 0,1 x 10-9 Reversible work for closed system define as ❑



W =−∫ PdV ❑



From isothermal compressibility k



k=



−1 dV T v dP



( )



Kita dapatkan



dV =−V kdP Lalu substitusikan dV dan W P2



W =V ∫ ( Vk ) dP P1 P2



W =V ∫ ( a−bP ) PdP P1



a b W =V ( P 2−P 1 )− (P2−P1) 2 3



[



❑



]



3,9 x 10−6 0,1 x 10−9 ( ) ( 30002−12) W =1 3000−1 − 2 3



[



W =16,65 ft 3 . atm



b.



Dik :



T = 298,15 K P = 12 bar Tc = 282,3 K Pc = 50,6 bar B = 140 cm3/mol C = 7200 cm6/mol R = 83,14 cm3 bar/mol K Tr = 1,056 K Pr = 0,2731 bar



a.



Z=



PV B C =1− − 2 RT V V



❑



]



RT −B P 83,14.298,15 V= −140 12 V =1926 cm3/mol V=



Iterasi RT B C V 1= (1− + 2 ) P Vo Vo 83,14.289,15 140 7200 V 1= (1− + ) 12 1926 19262 V 1=1919,834 140 7200 V 2=2066 1− + 1919,834 1919,834 2 V 2=1919,377



(



)



Setelah iterasi ke 5 diperoleh hasil V = 1919,34 cm3/mol Z = 0,8631



b.



Z=1+Bo



Pr Pr +ω B 1 Tr Tr



0,422 Tr 1,6 0,422 Bo=0,083− 1,056 1,6 Bo=−0,304 Bo=0,083−



0,172 Tr 4,2 B1=2,262 x 10−3 B1=0,139−



ω=0,089



Substitusikan ke persamaan Z Z=1+(−0,304) Z=0,9254 V=



ZRT P



0,2731 +¿ 1,056



0,9254.83,14 .298,15 12 V =1911,70028 cm3 /mol V=



c. Redlich/Kwong Equation Untuk persamaan RK: σεαΨβωΩ σ



=1



ε



=0



Ψ



= 0,42748



Ω



= 0,08664



α



= Tr-0,5 = 0,973124



q



=



β



=Ω



Z



= 1+ β−q β



Ψ .α Ω .Tr



= 5,070249



Pr = 0,022407 Tr Z −β (z +εβ )(Z + σβ)



Literasi Z = 1 Dan setelah literasi ke 4 di hasilkan Z



= 0,902665



V



=



V



= 1864,62 cm3/mol



Z .R.T P



d. Soave/Redlich/Kwong equation Untuk persamaan SRK : σ =1 ε



=0



Ψ



= 0,42748



Ω



= 0,08664



ω



= 0,089



α



= (1+(0,48508+1,5517.ω-0,1561.ω 2)(1-Tr0,5))2



α



= 0,965941



q



=



β



=Ω



Z



= 1+ β−q β



Ψ .α = 5,032823 Ω .Tr Pr = 0,022407 Tr Z −β (z +εβ )(Z + σβ)



Literasi Z = 1 Setelah literasi ke 3 dihasilkan : Z



= 0,903822



V



=



V



= 1867,009 cm3/mol



Z .R.T P



e. Peng/Robinson Equation Untuk persamaan RK : σ



= 2,41421



ε



= -0,41421



Ψ



= 0,45724



Ω



= 0,07779



α



= (1+(0,37464+1,54226.ω-0,2699.ω 2)(1-Tr0,5))2



α



= 0,97467



q



=



β



=Ω



Z



= 1+ β−q β



Ψ .α Ω .Tr



= 6,049815



Pr = 0,020118 Tr Z −β (z +εβ )(Z + σβ)



Literasi Z = 1 Setelah literasi ke 2 dihasilkan : Z



= 0,315212



V



=



V



= 651,1274 cm3/mol



Z .R.T P



3.



Diketahui : R



= 83,14 cm3.bar/mol.K



Untuk persamaan RK : σ



=1



ε



=0



Ψ



= 0,42748



Ω



= 0,08664



Mencari nilai Z liquid dan Vapour Zliquid = 1+ β−q β



Z −β (z +εβ )(Z + σβ)



Zvapour = β +( Z +εβ )(Z + σβ)



1+ β−Z qβ



Ditanya : V liquid & V vapour = ….?



a. Diketahui : α = Tr-0,5 = 1,087 q =



Ψ −3 /2 Tr = 6,332 Ω



β =Ω



Pr = 0,03302 Tr



di dapatkan hasilnya yaitu : Propan e



T (K) 313,15



Tc (K) 369,8



Tr (K) 0,847



P (bar) 13,71



Pc (bar) 42,48



Pr (bar) 0,323



α



Q



β



Zliquid



Vliquid



Zvapour



Vvapour



1,087



6,332



0,03302



0,824



1565,1



0,196



372,53



P (bar) 17,16



Pc (bar) 42,48



Pr (bar) 0,323



b. Diketahui α= Tr-0,5 = 1,070 q=



Ψ −3 /2 Tr = 6,040 Ω



β= Ω



Pr = 0,040 Tr



di dapatkan hasilnya yaitu : Propan e



T (K) 323,15



Tc (K) 369,8



Tr (K) 0,874



α



q



β



Zliquid



Vliquid



Zvapour



Vvapour



1,070



6,040



0,040



0,799



1250,22



0,212



323,3



c. Diketahui : α= Tr-0,5 = 1,067 q=



Ψ −3 /2 Tr = 5,999 Ω



β= Ω



Pr = 0,041 Tr



di dapatkan hasilnya yaitu : nbutane



T (K) 373,15



Tc (K) 425,1



Tr (K) 0,878



P (bar) 15,41



Pc (bar) 36,96



Pr (bar) 0,417



α



q



β



Zliquid



Vliquid



Zvapour



Vvapour



1,067



5,999



0,041



0,795



1599,86



0,215



432,23